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Consider the bivariate quadratic polynomial of the form: $$ x^2-by^2=z^k$$ where $k$ is an odd integer>2 and $b>0$. It's well-known that Euler's method: $$x^2-by^2=(p^2-bq^2)^k $$ provides a class of integer solutions. I am interested in finding a method that provides all the integer solutions not given by Euler's . I was reviewing the collection of algebraic identities by @Tito Piezas. He mentioned the following alternate method: $$x^2-by^2=(p^2-bq^2)^k(r^2-bs^2)$$ with $r^2-bs^2=1$. My question still remains, do these methods provide all the possible solutions?

  • $x^2+11y^2=3^3$ has a solution, but $p^2+11q^2=3$ does not. – Barry Cipra Jun 29 '16 at 14:45
  • Thanks for pointing that out. Clearly, I meant $k>1$. Btw, $p^2+11q^2=3$ has solutions. They just are not integers. –  Jun 29 '16 at 14:49
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    My comment does use a $k\gt1$. To rephrase it, how do you obtain the solution $(x,y,z)=(4,1,3)$ to the equation $x^2+11y^2=z^3$ from solutions to the equation $x^2+11y^2=(p^2+11q^2)^3$? (And yes, I was clearly referring to integer solutions. Along the same lines, it's amusing to note that the equation $x^n+y^n=z^n$ can have positive integer solutions $(x,y,z)$ for exponents $n\gt2$, just not integer exponents.) – Barry Cipra Jun 29 '16 at 15:05
  • so I guess your point is Euler's method does not provide all the solutions. So I am asking you how do I obtain them all? If you have any further insight, please enlighten me. –  Jun 29 '16 at 16:08
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    Sorry, I don't have any further insight. Hopefully someone else will. You might want to edit the question to acknowledge that Euler's method provides some but not necessarily all integer solutions, so the real question is how can you get the rest. If nothing else, it'll help bring the question back to the attention of people who might know the answer. – Barry Cipra Jun 29 '16 at 17:06

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Dickson’s complete solution for the equation $x^2-my^2=zw$ should give all solutions.

Kieren MacMillan
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  • Thanks. Just a question, can smaller factors of $Z$ and $W$ written the same way? I mean $ea^2+2fab+gb^2$ – NumThcurious Oct 17 '18 at 09:13
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    @NumThcurious: Of course! If $x^2 - my^2 = zw$, then also $x^2-my^2= z'w'$ with integers $z'=z/a$ and $w'=aw$, where $a$ is some factor of $z$ [and similar for factors of $w$]. – Kieren MacMillan Oct 18 '18 at 13:39
  • Did you mean $$x^2-my^2=a^2z'w'$$? Assuming that all the smaller primes of $Z,W$ are of the form $$eu^2+2fuv+gv^2$$ Wouldn't that imply if $$x^2-my^2=(MN)^p$$ then $M,N$ are also of the same form? – NumThcurious Oct 20 '18 at 01:35
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    @NumThcurious: No, I didn’t mean $a^2$. All I mean(t) was that any possible way that you factor the right-hand side must result in some $z$ and $w$, which are then subject to Dickson’s parameterization. For example, say $x^2-my^2=12$; then you have $(z,w) \in {(1,12), (2,6), (3,4), (4,3), (6,2), (12,1)}$ [and the same with negative components], and each of those factor-pairs (“smaller factors of $Z$ and $W$”, in your original wording) must satisfy Dickson’s parameterization. – Kieren MacMillan Oct 20 '18 at 14:24
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    @NumThcurious: As to your second question… Yes, that’s the implication. – Kieren MacMillan Oct 20 '18 at 14:25
  • That makes sense. Thank you very much. – NumThcurious Oct 21 '18 at 15:35
  • x^2-76y^2=-1342368 = -2^5×3^2×59×79 (9 prime factors, 4 distinct) – NumThcurious Oct 21 '18 at 19:42
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    @NumThcurious: The divisors of $-1342368$ are ${1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 59, 72, 79, 96, 118, 144, 158, 177, 236, 237, 288, 316, 354, 472, 474, 531, 632, 708, 711, 944, 948, 1062, 1264, 1416, 1422, 1888, 1896, 2124, 2528, 2832, 2844, 3792, 4248, 4661, 5664, 5688, 7584, 8496, 9322, 11376, 13983, 16992, 18644, 22752, 27966, 37288, 41949, 55932, 74576, 83898, 111864, 149152, 167796, 223728, 335592, 447456, 671184, 1342368}$, with appropriate sign(s). Each of these should have a quadratic representation with the same discriminant (as given by $m=76$). – Kieren MacMillan Oct 21 '18 at 19:58
  • I was in the process of writing a comment but did not finish. I did not notice that it was actually posted. – NumThcurious Oct 21 '18 at 20:08
  • I was about to post the following: $$x^2-76y^2=(5l^2+22lq+9q^2)(5n^2-22nq+9r^2)$$ $$=-1342368 = -2^5×3^2×59×79$$ (9 prime factors, 4 distinct) where $$x=\pm 436$$ $$y=\pm 142$$ $$l=3$$ $$q=7$$ $$n=13$$ $$r=17$$ However there exist no couple $(a,b)$ such that $2,3,59,79$ can be written as $5a^2+22ab+9b^2$. I am not sure where I went wrong. – NumThcurious Oct 21 '18 at 20:09
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    I see where I went wrong. The discriminant $m$ must be $76$ . However, $(e,f,g)$ will take different values other than just $$(5,11,9)$$ depending of the prime chosen. – NumThcurious Oct 21 '18 at 20:24